The integral
Question:

The integral $\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x$ is equal to

(where $C$ is a constant of integration):

  1. (1) $\tan x-\frac{x \sec x}{x \sin x+\cos x}+C$

  2. (2) $\sec x+\frac{x \tan x}{x \sin x+\cos x}+C$

  3. (3) $\sec x-\frac{x \tan x}{x \sin x+\cos x}+C$

  4. (4) $\tan x+\frac{x \sec x}{x \sin x+\cos x}+C$


Correct Option: 1

Solution:

$\int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$

$\because \frac{d}{d x}(x \sin x+\cos x)=x \cos x$

$=\int \frac{x \cos x}{(x \sin x+\cos x)^{2}}\left(\frac{x}{\cos x}\right) d x$

$=\frac{x}{\cos x}\left[\frac{-1}{x \sin x+\cos x}\right]$

$-\int \frac{x \sin x+\cos x}{\cos ^{2} x}\left[\frac{-1}{x \sin x+\cos x}\right] d x$

$=\frac{x}{\cos x}\left[\frac{-1}{x \sin x+\cos x}\right]+\int \sec ^{2} x d x$

$=\frac{-x \sec x}{x \sin x+\cos x}+\tan x+C$

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